If $\{v_1,v_2,v_3,...,v_n\}$ is a set of linearly dependent vectors then there exists atleast one $k_i:i=1,2,3,...,n$ such that $k_1v_1+k_2v_2+k_3v_3+...+k_nv_n=0$

Consider $k_1T(v_1)+k_2T(v_2)+k_3T(v_3)=0$

A linear transformation T: U→V satisfies the following properties:

1. $T(u_1+u_2)=T(u_1)+T(u_2)$

2. $T(au)=aT(u)$

Here, $u,u_1,u_2$ ∈ U

As T is a linear transformation,

$k_1T(v_1)+k_2T(v_2)+k_3T(v_3)=0\\T(k_1v_1)+T(k_2v_2)+T(k_3v_3)=0\\T(k_1v_1+k_2v_2+k_3v_3)=0\\$

As $\{v_1,v_2,v_3\}$ is a linearly dependent set,

$k_1v_1+k_2v_2+k_3v_3=0$ for some $k_i\neq 0:i=1,2,3$

So, for some $k_i\neq 0:i=1,2,3$

$k_1T(v_1)+k_2T(v_2)+k_3T(v_3)=0$

Therefore, set $\{T(v_1), T(v_2), T(v_3)\}$ is linearly dependent.

6 0

3 years ago

BRAINLIEST ANSWER UP FOR GRABS!!

Svetach [21]

X = sqrt(5^2 - 3^2) = 4
y = sqrt(7^2 - 3^2) = sqrt 40

Horizontal distance = x + y + 3 = 7 + sqrt40

8 0

3 years ago

Read 2 more answers

I need help with this qustion ?

emmainna [20.7K]

The answer should be 32.

4 0

3 years ago